with $0 \leq \theta \leq \pi$ and $0 \leq \phi < 2 \pi$. Notice that if you have a complex coefficient for $|0\rangle$, you have to factor it and forget about the global phase of the qbit.

As the basis ($| 0 \rangle$, $|1\rangle$) is orthonormal, we have $\langle 1 | 0 \rangle = 0$, and we can identify the coefficients of the decomposition. You have

$|\psi\rangle = (1/\sqrt{2}) |0\rangle + i (1/\sqrt{2}) |1\rangle $

so

$\cos(\theta/2) = \frac{1}{\sqrt{2}}$

and thus (with inverse functions, if you want) $\theta=\pi/2$. So we have also

$\sin(\theta/2) = \frac{1}{\sqrt{2}}$

and we must have $\sin(\phi) = 1$ and $\cos(\phi) = 0$, which gives $\phi=\pi/2$ too.

After some time, you will see where the qbit goes on the Bloch sphere without tedious calculations. For example, here you can see that the coefficients of $| 0 \rangle$ and $|1\rangle$ have the same magnitude (it happens when $\theta=\pi/2$), so the state lies on the equator. Now, you can see that the $\phi$ coordinate is just the cylindrical coordinate on the circle. It means that the relative phase between $|0\rangle$ and $|1\rangle$ gives the angular position of the qbit on the $xy$ plane. Conversely, the relative magnitude between $|0\rangle$ and $|1\rangle$ gives you the angular position in the plane which contains $z$ and your state.

You can obtain the coordinates in 3-space corresponding to the Bloch sphere, by taking expectation values of the Pauli spin operators:
$$\begin{align*} x &= \langle\psi | \sigma_x |\psi\rangle \\ y &= \langle\psi | \sigma_y |\psi\rangle \\ z &= \langle\psi | \sigma_z |\psi\rangle \end {align*}$$
and in the case of the state you describe, we have $(x,y,z) = (0,1,0)$, i.e. your state is the one on the positive $y$ axis.